0
$\begingroup$

I'm having trouble computing the integral $\int_{A}\log(\sin(x-y))dx dy$, where $A$ is the triangle defined by the 3 eqations: $x = \pi, y = x ,y = 0$

I tried the substition $u = x-y, v = x+y$ and got:

$$\frac{-1}{2}\int_{0}^{2\pi}\int_{0}^{v}\log(\sin u))du dv$$

I'm quite new at this, so there's a possibility I've made a silly mistake, or I'm just not getting something obvious about how to continue.

Any help would be appreciated!

$\endgroup$
1
$\begingroup$

The integral you are dealing with is: $$\int_0^{\pi} \int_0^x \ln(\sin(x-y))\,dy\,dx=\int_0^{\pi} \int_0^x \ln(\sin y)\,dy\,dx$$ Changing the order of integration: $$\int_0^{\pi} \int_y^{\pi} \ln(\sin y)\,dx\,dy=\int_0^{\pi} (\pi-y)\ln(\sin y)\,dy=I$$ $I$ is equivalent to: $$I=\int_0^{\pi}y\ln(\sin y)\,dy$$ Add the two expressions for $I$ and you get: $$I=\frac{\pi}{2}\int_0^{\pi} \ln(\sin y)\,dy=\boxed{-\dfrac{\pi^2}{2}\ln 2}$$

$\endgroup$
  • $\begingroup$ Thanks for the answer. How is the first equation justified? i.e., why does $sin(x-y)$ become $sin(y)$? Also, what allows you to change the order of integration? I thought that Fubini's theroem only allows this in the case of a rectangle domain. $\endgroup$ – Robert Jun 21 '14 at 10:49
  • 1
    $\begingroup$ @Robert: I used the property of definite integrals which states that: $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ About your second question, I am not sure what you ask. Do you mean changing the order of integration is invalid here? I am not sure about the correct terminology in double integrals. $\endgroup$ – Pranav Arora Jun 21 '14 at 10:58
  • $\begingroup$ Why is it possible to integrate by x and then by y instead of by y and then by x? I know that in that case of a rectangle domain (not a triangle like here) it is possible to switch the order of integration by Fubini's theorem. Why is it justified when the domain is a triangle? $\endgroup$ – Robert Jun 21 '14 at 11:04
  • $\begingroup$ @Robert: I didn't know if there was a theorem governing the change of order of integration in double integrals. I did it based on my intuition. I am sorry, I can't help with that. :( $\endgroup$ – Pranav Arora Jun 21 '14 at 11:07
  • $\begingroup$ You're probably correct anyway, thanks for the answer! $\endgroup$ – Robert Jun 21 '14 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.